ABCD is a square. E is a point on the side BC such that BE = BC/3, and F is a point on the ray DC such that CF = DC/2. Prove that the lines AE and BF intersect on the circumcircle of the square.
Solution
Let X be the point of intersection. tan BEA = 3, tan EBX = 1/2, so tan BXE = tan(BEA - EBX) = (3 - ½)/(1 + 3/2) = 1. Hence ∠BXE = 45o. So ∠AXB = ∠ACB. Hence X lies on the circle.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03